Solve the cube root equation
$$\sqrt[3]{x+28} - \sqrt[3]{x-28} = 2 $$

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Uniteasy Admin · Accepted Answer

Let		
	$$a = \sqrt[3]{x+28} $$nc	
	$$b = \sqrt[3]{x-28} $$nc	
Then		
	$$a - b = 2$$nc	
Difference of $$a$$ and $$b$$ raised to the power of 3 gives		
	$$a^3 - b^3 = 56$$nc	
Factor LHS using difference of cubes formula		
	$$(a - b)(a^2+ab+b^2) = 56$$nc	
Substitute (3) to (5) and simplify 		
	$$a^2+ab+b^2 = 28$$nc	
Square of equation (3) gives		
	$$(a -b )^2 = 4$$c	
Expand LHS		
	$$a^2 - 2ab +b^2 = 4$$nc	
Substract (7) from (6) and solve for $$ab$$		
	$$ ab = \dfrac{1}{3}(\dfrac{2\cdot 28}{2} - 2^2 ) = 8$$nc	
Substituting $$ab$$ to (6) gives		
	$$a^2+b^2 = 28 - ab = 20$$nc	
Construct a perfect square binomial with (9) and (8)		
	$$(a+b)^2 = a^2+b^2+2ab$$c	
	$$(a+b)^2 = 20 +2\cdot 8 = 36$$c	
Taking square root of both sides gives		
	$$a +b = \pm6$$nc	
Now we get two systems of equations		
Case 1		
	$$\begin{cases} a+b = 6 \ a - b =2 \end{cases}$$c	
which gives		
	$$\begin{cases} a= 4 \ b =5 \end{cases}$$c	
Then		
	$$x = a^3 - 28 = 4^3 - 28 = 36$$c	
Case 2		
	$$\begin{cases} a+b = -6 \ a - b =2 \end{cases}$$c	
which gives		
	$$\begin{cases} a= -2 \ b =-4 \end{cases}$$c	
Then		
	$$x = a^3 - 28 = -2^3 - 28 = -36$$c	
In summary, we get two solutions for $$x$$.		
	$$x = 36 	ext{ or }x = -36$$c

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